If a is a rational number such that `(x-a)` is a factor of `x^(3)-3x^(2)-3x +9`, then

Options

A. a may be any integer
B. a is an integer divisible by 9
C. a cannot be integer
D. a can have three values

Answer

To determine the rational number \( a \) such that \( (x-a) \) is a factor of the polynomial \( x^3 - 3x^2 - 3x + 9 \), we will follow these steps:

### Step 1: Factor the polynomial
We start with the polynomial:
\[ x^3 - 3x^2 - 3x + 9 \]

To factor this polynomial, we can use the Rational Root Theorem or synthetic division to find potential rational roots. We will test \( x = 3 \) as a possible root.

### Step 2: Synthetic Division
We perform synthetic division of \( x^3 - 3x^2 - 3x + 9 \) by \( x - 3 \):

```
3 | 1 -3 -3 9
| 3 0 -9
-------------------
1 0 -3 0
```

The result is \( x^2 - 3 \) with a remainder of 0, indicating that \( x - 3 \) is indeed a factor.

### Step 3: Factor the remaining polynomial
Now we have:
\[ x^3 - 3x^2 - 3x + 9 = (x - 3)(x^2 - 3) \]

Next, we can factor \( x^2 - 3 \) further:
\[ x^2 - 3 = (x - \sqrt{3})(x + \sqrt{3}) \]

### Step 4: Write the complete factorization
Thus, the complete factorization of the polynomial is:
\[ x^3 - 3x^2 - 3x + 9 = (x - 3)(x - \sqrt{3})(x + \sqrt{3}) \]

### Step 5: Identify the values of \( a \)
From the factorization, we can see that the factors are \( x - 3 \), \( x - \sqrt{3} \), and \( x + \sqrt{3} \). Therefore, the values of \( a \) can be:
\[ a = 3, \quad a = \sqrt{3}, \quad a = -\sqrt{3} \]

### Conclusion
The rational number \( a \) such that \( (x-a) \) is a factor of the polynomial \( x^3 - 3x^2 - 3x + 9 \) can take three values: \( 3, \sqrt{3}, -\sqrt{3} \). Among these, only \( 3 \) is a rational integer.